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4 years ago

−8y2−9y)+(8y3+9y2−2y)=

Mathematics

2 answers:

aalyn [17]4 years ago

6 0

8 y3+y2-11y would be the simplified answer.

Alik [6]4 years ago

3 0

Combine like terms:

(-8y² - 9y) + (8y³ + 9y² - 2y)

8y³ - 8y² + 9y² - 9y - 2y

8y³ (- 8y² + 9y²) (- 9y - 2y)

8y³ + y² - 11y

8y³ + y² - 11y is your answer

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Any real number line ranges from negative infinity to positive infinity. A real number number line consists of all the rational and irrational numbers. Let us take three intervals which contain both the rational and irrational numbers.

First interval: [3,4]

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Second interval : [0,2]

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Third interval : [2,3]

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3 years ago

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4 years ago

Which point best approximates ? A B C D

snow_lady [41]

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6 0

3 years ago

Read 2 more answers

Find an equation of the circle that satisfies the given conditions.

hodyreva [135]

The equation of a circle:
$(x-h)^2+(y-k)^2=r^2$
(h,k) - the coordinates of the centre
r - the radius

The midpoint of the diameter is the centre of a circle.
The coordinates of the midpoint:
$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
(x₁,y₁), (x₂,y₂) - the coordinates of endpoints

$P(-1,1) \\
x_1=-1 \\ y_1=1 \\ \\ Q(5,9) \\ x_2=5 \\ y_2=9 \\ \\
\frac{x_1+x_2}{2}=\frac{-1+5}{2}=\frac{4}{2}=2 \\ \frac{y_1+y_2}{2}=\frac{1+9}{2}=\frac{10}{2}=5$

The centre of the circle is (2,5).

The radius is the distance between an endpoint of the diameter and the centre.
The formula for distance:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(-1,1) \\ x_1=-1 \\ y_1=1 \\ \\ (2,5) \\ x_2=2 \\ y_2=5 \\ \\ d=\sqrt{(2-(-1))^2+(5-1)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

The radius is 5.

$(x-2)^2+(y-5)^2=5^2 \\
\boxed{(x-2)^2+(y-5)^2=25}$

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Answer:

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